Question:
If the maximum kinetic energy of emitted photo electrons from a metal is 0.9 eV and work function is 2.2 eV then the energy and wavelength of incident radiation are
If the maximum kinetic energy of emitted photo electrons from a metal is 0.9 eV and work function is 2.2 eV then the energy and wavelength of incident radiation are
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Incident Energy = Work Function + Kinetic Energy. Use 12400 for quick eV-to-Angstrom conversions.
Updated On: Jul 4, 2026
- 3.1 eV, 4000 \AA
- 2.2 eV, 2000 \AA
- 2.2 eV, 4000 \AA
- 3.1 eV, 2000 \AA
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Einstein's photoelectric equation states: $E = \Phi + K_{max}$, where $E$ is photon energy, $\Phi$ is work function, and $K_{max}$ is maximum kinetic energy.
Step 2: Meaning
Wavelength ($\lambda$) is related to energy ($E$) in eV by the approximate formula $\lambda (\text{\AA}) = 12400 / E(\text{eV})$.
Step 3: Analysis
Incident Energy $E = 2.2 \text{ eV} + 0.9 \text{ eV} = 3.1 \text{ eV}$. Wavelength $\lambda = 12400 / 3.1 \approx 4000 \text{\AA}$.
Step 4: Conclusion
The energy is 3.1 eV and the corresponding wavelength is 4000 \AA. Final Answer: (A)
Einstein's photoelectric equation states: $E = \Phi + K_{max}$, where $E$ is photon energy, $\Phi$ is work function, and $K_{max}$ is maximum kinetic energy.
Step 2: Meaning
Wavelength ($\lambda$) is related to energy ($E$) in eV by the approximate formula $\lambda (\text{\AA}) = 12400 / E(\text{eV})$.
Step 3: Analysis
Incident Energy $E = 2.2 \text{ eV} + 0.9 \text{ eV} = 3.1 \text{ eV}$. Wavelength $\lambda = 12400 / 3.1 \approx 4000 \text{\AA}$.
Step 4: Conclusion
The energy is 3.1 eV and the corresponding wavelength is 4000 \AA. Final Answer: (A)
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